$\mathbb{Z}_pH$-module

Solution 1:

$\renewcommand{\phi}{\varphi}$$\newcommand{\Z}{\mathbb{Z}}$$\DeclareMathOperator{\End}{End}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$In general, you cannot expect $V$ to be a faithful $\mathbb{Z}_{p} H$-module.

Let $G = A_{4}$, and $V$ be its unique Sylow $2$-subgroup. Let $H = \Span{a}$ be a Sylow $3$-subgroup of $G$. Now $H$ acts faithfully on $V$, but $\mathbb{Z}_{2} H$ does not, as $1 + a + a^{2}$ is in the annihilator.

Addendum

Somebody more versed than me in these matters will be able to put what I am about to say in proper context.

In the general setting of OP, let $\phi : \Z_{p} H \to \End(V)$ be the homomorphism defining the $\Z_{p} H$-module structure on $V$.

Suppose there is an element $h \in H$ of order $n$ such that the minimal polynomial $m \in \Z_{p}[x]$ of $\phi(h)$ over $\Z_{p}$ is a proper divisor of $x^{n} - 1$. (This will be the case if $\dim(V) < n$, as in the $A_{4}$ example, because $m$ divides the characteristic polynomial of $\phi(h)$, which has degree $\dim(V)$.)

If the degree of $m$ is $d < n$, the elements $1, h, \dots, h^{d}$ are distinct, thus independent in $\Z_{p} H$. It follows that $0 \ne m(h) \in \Z_{p} H$ is an element in the kernel of $\phi$, as $\phi(m(h)) = m(\phi(h)) = 0$, so that the module is not faithful.